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4.7
Optimization Problems
Steps to follow for max and min problems.
(a) Draw a diagram, if possible.
(b) Assign symbols to unknown quantities.
(c) Assign a symbol,Q, to the quantity to
be maximized or minimized.
(d) Express Q in terms of the other symbols.
(e) Eliminate all but one unknown symbol,
say x, from Q.
(f) Find the absolute maximum or minimum
of Q = f (x).
 i  State the domain of f  x  .
 ii  Find the critical number of f  x  .
(iii) Use the first derivative
test to determine whether f  x  has a
max or a min at the critical number.
(iv) Does the absolute max or min
value of f  x  occur at an end point?
1. Example: Find two numbers whose
difference is 100 and whose product
is a minimum.
Solution:Let x denote one of the numbers
and y the other.
x  y  100 or y  x  100.
Put P  xy
=˃p(x)=x(x-100)
(i) Domain of P(x): all real numbers.
(ii) Critical numbers of P(x)
P′(x)= 1 x  100   x 1
 2 x  100
 0  x = 50.
(iii) First Derivative Test


P′(x)
50
P( x)
50
P(50) is an absolute min value.
(iv) When x  50,
50  y  100  y  50.
 x  50, y  50.
Graph:
P( x)  x( x  100)
2. Example: A farmer wants to fence a
rectangular enclosure for his horses
and then divide it into thirds with fences
parallel to one side of the rectangle. If
he has 2000m of fencing, find the area
of the largest rectangle that can be
enclosed.
Solution: Let l be the length and w the
width of the rectangle.
w
l
A  lw
P  4w  2l  2000
2000  2l
w
4
2000  2l 

 A  l  l
4
2
1000  l 

l
 500l 
l
2
2
(i) Domain of A(l): 0  l  1000.
(ii) Critical numbers of A(l)
A  l   500  l
 A  l   0
 l  500.
By the first derivative test we
can show that A(500) is the
max value of A,
When l=500,
2000  2l
w
4
2000  1000

 250.
4
max area of the rectangle = 500  250
 125,000 meters
2
2
Graph:
l
A(l )  500l 
2
3. Example:Find the point on the parabola
x=y2 that is closest to the poit (0,3).
Let D denote the distance from (0,3) to
any point (x,y) on the parabola.
D  ( x  0)  ( y  3)
2
 ( y )  ( y  3)
2 2
D(y)=
y  ( y  3)
4
2
2
2
Find the value of y which makes D(y)
a minimum.
(i) Domain of D(y): all real numbers > 0.
(ii) Critical numbers:
D′(y)

2y  y  3
3
y  ( y  3)
4
2
( y  1)(2 y  2 y  3)
2
y  ( y  3)
2
 0  y  1, since 2 y  2 y  3  0
for all y.
4
2
(iii) First Derivative Test
D′(y)


1
By the first derivative test D(y)
has an absolute minimum when y  1.
When y  1, x  y  1.
2
 1,1 is the required point
2
on the graph of x  y .
Graph:
D( y ) 
y  ( y  3)
4
2
4. Example: A right circular cylinder is
enscribed in a sphere of radius r. Find the
largest possible volume of such a cylinder.
Solution:
Let x be the radius,
2 y be the height and
V the volume of the
cylinder.
V   x (2 y ).
2
r
y
x
V  x
x y r
2
2
2
2 y
2
x r y
2
2

2
 V( y )   r  y
2
2
2y
(i) Domain of V(y): 0  y  r
(ii) Critical numbers of V(y)
V( y)  2π r2 -3y2
2
2
V( y)  0  3 y  r
r
 y
.
3
By the first derivative test
r
That is V  y  is a maximum where y 
.
3
Note that V  0   V  r   0.
Maximum value for V( y ) is
3

 r 
 r 
2 r
V



 2  r

3  3  
 3

3 
3
3
 r3

r
3r  r
 2 

  2 

 3 3 3
 3 3 
3
3
2r
4

r
 2

.
3 3
3 3
5. Example: Mountain Beer sells its beer
in an aluminum can in the shape of a rightcircular cylinder. The volume of each can
is 250 ml. What should the dimension of
the can be in order to minimize the amount
of aluminum used?
Solution:
Let h be the height, r the radius, V the
volume and S the surface area of the can.
2
S  2 r h  2 r and V   r 2 h  250
250  250
 h

2
2
r
r
2 r
r
h
Surface
Area,
S  2 r h   r  r
2
 2 r h  2 r
Volume  Area of base  height
2
2
  r  h   r h.
2
2
S  2 r h  2 r
2
500
S(r)= 2 r  2 r 2  2 r 
r
r
(i) Domain of S(r) r  0
2
250
(ii) Critical numbers of S(r)
500
S′(r)= 4 r  2
r
3
4 r  500 

2
r
2
r  125  5cm
S′(r)=0 =˃
(iii) First Derivative Test 3
4 r  125
S′(r)=
2
r
S′(r)
5
0
3


S(r)


5
0
S(r) has its absolute minimum value
when r  5cm.
when r  5, h 
250
2
r
250

25
 10cm
 The surface area is a minimum
when r  5cm and h  10cm.
Graph:
500
S (r )  2 r 
r
2